Question: $f(x) = \dfrac{ 9 }{ \sqrt{ 1 - \lvert x \rvert } }$ What is the domain of the real-valued function $f(x)$ ?
Explanation: First, we need to consider that $f(x)$ is undefined anywhere where the radicand (the expression under the radical) is less than zero. So we know that $1 - \lvert x \rvert \geq 0$ This means $\lvert x \rvert \leq 1$ , which means $-1 \leq x \leq 1$ Next, we need to consider that $f(x)$ is also undefined anywhere where the denominator is zero. So we know that $\sqrt{ 1 - \lvert x \rvert } \neq 0$ , so $\lvert x \rvert \neq 1$ This means that $x \neq 1$ and $x \neq -1$ So we have four restrictions: $x \geq -1$ $x \leq 1$ $x \neq -1$ , and $x \neq 1$ Combining these four, we know that $x > -1$ and $x < 1$ ; alternatively, that $-1 < x < 1$ Expressing this mathematically, the domain is $\{ \, x \in \RR \mid -1< x <1\, \}$.